Let $X$ be a separable Banach space. Let $B_1(X)$ be the set of all bounded linear operators $X \to X$ with operator norm $\leq 1$. Does $B_1(X)$ have to be separable in the strong operator topology?
Separability of unit ball in strong operator topology
banach-spacesfunctional-analysisoperator-theory
Related Solutions
The answer is in part I of Linear Operators by Dunford and Schwartz, Theorem VI.1.4. For any Banach spaces $X$ and $Y$ linear functionals on $\mathcal{B}(X,Y)$ continuous in strong and weak operator topologies are the same. It follows from the proof that they are of the form $F(A)=\sum_{i=1}^n\varphi_i(Ax_i)$ for $x_i\in X$ and $\varphi_i\in Y^*$. In other words, the dual space can be identified with the algebraic tensor product $X\otimes Y^*$, or equivalently the space of finite rank operators $\mathcal{F}(Y,X)$.
For the ultraweak and the ultrastrong topologies on $\mathcal{B}(H)$, where $H$ is a Hilbert space, the answer is in Von Neumann Algebras by Dixmier, Lemma I.3.2. Again, the dual spaces coincide and can be identified with the projective tensor product $H\otimes_\pi H$, or the space of the trace class operators $\mathcal{L}_1(H)$. It also happens to coincide with the norm closure of $\mathcal{F}(H)$ in the Banach dual $\mathcal{B}(H)^*$, and is a Banach predual $\mathcal{B}(H)_*$ of $\mathcal{B}(H)$. Moreover, the ultraweak topology on $\mathcal{B}(H)$ coincides with the weak* topology generated by this predual, they are both $\sigma\big(\mathcal{B}(H),H\otimes_\pi H\big)$.
It doesn't appear that the ultraweak or the ultrastrong topologies were studied much beyond Hilbert spaces, I couldn't even find any established definitions for them. Since the weak operator topology coincides with $\sigma\big(\mathcal{B}(X,Y), X\otimes Y^*\big)$, by analogy to Hilbert spaces a natural candidate for the ultraweak is $\sigma\big(\mathcal{B}(X,Y), X\otimes_\pi Y^*\big)$. However, according to Introduction to Tensor Products of Banach Spaces by Ryan, Section 2.2 a predual to $\mathcal{B}(X,Y)$ is $X\otimes_\pi Y_*$, where $Y_*$ is a predual to $Y$. So the ultraweak so defined does not coincide with the weak* unless $Y$ is reflexive. And even if $Y$ is reflexive it is not clear what a natural candidate for the ultrastrong is or how the duals are related.
We want to show that a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 1 if and only if a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 2.
Suppose that $T_{\lambda}\to T$ with respect to $\textbf{def}$ 1, then we know that $T_{\lambda}x\to Tx$ for each $x\in X$ meaning that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T).$$ Putting this in context, for each $F_x\in \{F_x:\beta(X,Y)\to Y \text{ such that }F_x(T)=Tx, T\in\beta(X,Y)\}$ we have $F_x(T_{\lambda})\to F_x(T)$. So $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2.
On the other hand, if $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2, then we know that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T),$$ for each $x\in X$ and as a result $T_{\lambda}x\to Tx$ for each $x\in X$ and we have convergence with $\textbf{def}$ 1. It is simply unpacking the definitions.
Best Answer
It turns out the answer is yes. Let $(x_k)_{k \geq 1}$ be a countable dense subset of $X$ and consider the set $A = \{ (Tx_k)_{k \geq 1} : T \in B_1(X) \} \subseteq X^{\mathbb{N}}$. Since $X$ is separable and metrizable, $X^{\mathbb{N}}$ is also separable and metrizable, and therefore $A$ is also separable. Let $(T_n)_{n \geq 1}$ be an enumeration of the operators corresponding to a countable dense subset of $A$.
The set $\{T_n\}_{n \geq 1}$ is dense in $B_1(X)$ for the strong operator topology. To see this, fix $T \in B_1(X)$ and consider the sequence $(Tx_k)_{k \geq 1} \in A$. By construction there is a subsequence $(T_{n_j})_{j \geq 1}$ such that $(T_{n_j} x_k)_{k \geq 1} \to (Tx_k)_{k \geq 1}$ in $A$ as $j \to \infty$. In particular this implies that $T_{n_j} x_k \to T x_k$ as $j \to \infty$ for each fixed $k$. Finally because $\{x_k\}_{k \geq 1}$ is dense in $X$ this implies that $T_{n_j} \to T$ in the strong operator topology.